Double integrals - polar coordinates - how do I determine the boundaries for $r$ and $\phi$?

Question

I need to calculate the surface bounded by:

1. $(x-1)^2 +y^2=1$

2. $y=x$

3. $y=0$

and I have to do it using polar coordinates. Now, letting $x=r\cos{\phi}$ and $y=r\sin{\phi}$, $\phi \in [0, 2\pi]$, $r \in [0, \infty)$, we have:

1. $ 0 \le r \le 2\cos{\phi}$

This is obvious because the area needs to be inside of the circle

2. $r\sin{\phi} = 0$

3. $\sin\phi = \cos\phi$

I know that the boundaries when we integrate by $dr$ are $0$ and $2\cos\phi$, but how can I deduce the boundaries for $\phi$?

The "equals" sign doesn't help in $2.$ and $3.$, because, for example, in $2$, both $r$ and $\sin\phi$ can be zero.

Can anyone help?

Answer 1

Picture

$(x-1)^2+y^2=1$ is a circle, $y=x$ is the slanted line, $y=0$ is the horizontal line.

The two lines are $\theta = \pi/4$ and $\theta = 0$. We look a the picture to see that the values we want for $\theta$ are from $0$ to $\pi/4$. The area we want is $0 \le \theta \le \pi/4, 0 \le r \le 2\cos\theta$.
The double integral for the area is $$ \int_0^{\pi/4}\int_0^{2\cos\theta} r\;d r\;d\theta $$

We have to remember that $dx\,dy$ becomes $r\,dr\,d\theta$ because the Jacobian for the change of variables is $r$.